Solving for Maximum Spring Length With a Dropped Clay Ball

[maki]

A massless pan hangs from a spring that is suspended from the ceiling. When empty, the pan is 50cm below the ceiling. If a 100g clay ball is placed gently on the pan, the pan hangs 60cm below the ceiling. Suppose the clay ball is dropped from the ceiling onto an empty pan. What is the pan's distance from the ceiling when the spring reaches its maximum length?

I figured I'd start by finding the final velocity of the ball using the gravitational potential energy.
Ugi=mgy
Ugi=(0.100kg)(9.8m/s)(0.5m)
Ugi=0.49 J
Ugf=0 J
Vf=sqrt[(-2 * (Ugf - Ugi)) / m]
Vf=sqrt[(-2 * -0.49) / 0.5)]
Vf=sqrt(1.96)
Vf=1.4 m/s

Next, I found the spring constant (k) using the stationary weight of the ball in the pan.
Fsp=mg
Fsp=(0.1)(9.8)=.98 N
Fsp=k*(Delta S, or change in spring length)
.98 N=k*0.1
k=9.8

Now I'm stuck, can't seem to find the right formula to use to get any farther, I should have just the one more step I think. If someone could post a formula or explain where I need to go next I would appreciate it. I have a lab in 15 minutes, but I'll be back after that to work on this some more.

Thanks (=

 

[Päällikkö]

Potential energy of a spring:
W = \frac{1}{2}kx^2